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IB Mathematics: Applications and Interpretation HL · AHL 6: Complex Numbers · AHL 6.10

Adding sinusoids Problem set A One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. Write \(9\cos(4t + 0.3) + 5\cos(4t + 1.9)\) in the form \(A\cos(4t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

    • \(9\cos(4t + 0.3) + 5\cos(4t + 1.9) = \mathrm{Re}\left(9e^{(4t + 0.3)i} + 5e^{(4t + 1.9)i}\right)\)
    • \(= \mathrm{Re}\left(e^{4ti}(9e^{0.3i} + 5e^{1.9i})\right)\)
    • From the calculator: \(9e^{0.3i} + 5e^{1.9i} = 6.98 + 7.39i = 10.2e^{0.814i}\)
    • \(= \mathrm{Re}\left(10.2e^{(4t + 0.814)i}\right) = 10.2\cos(4t + 0.814)\)

  2. Write \(11\sin(t + 0.7) + 6\sin(t + 2.9)\) in the form \(A\sin(t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

    • \(11\sin(t + 0.7) + 6\sin(t + 2.9) = \mathrm{Im}\left(11e^{(t + 0.7)i} + 6e^{(t + 2.9)i}\right)\)
    • \(= \mathrm{Im}\left(e^{ti}(11e^{0.7i} + 6e^{2.9i})\right)\)
    • From the calculator: \(11e^{0.7i} + 6e^{2.9i} = 2.59 + 8.52i = 8.91e^{1.28i}\)
    • \(= \mathrm{Im}\left(8.91e^{(t + 1.28)i}\right) = 8.91\sin(t + 1.28)\)

  3. Use complex exponentials.

    1. (a) Show that \(6\sin(3t + 2) + 2\sin(3t + 0.5) = \mathrm{Im}\left(6e^{(3t + 2)i} + 2e^{(3t + 0.5)i}\right)\).

      • \(e^{\theta i} = \cos\theta + i\sin\theta \text{, so } \mathrm{Im}\left(Ae^{\theta i}\right) = A\sin\theta\)
      • \(\mathrm{Im}(z_1 + z_2) = \mathrm{Im}(z_1) + \mathrm{Im}(z_2) = 6\sin(3t + 2) + 2\sin(3t + 0.5)\)

    2. (b) Hence write \(6\sin(3t + 2) + 2\sin(3t + 0.5)\) in the form \(A\sin(3t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(6\sin(3t + 2) + 2\sin(3t + 0.5) = \mathrm{Im}\left(6e^{(3t + 2)i} + 2e^{(3t + 0.5)i}\right)\)
      • \(= \mathrm{Im}\left(e^{3ti}(6e^{2i} + 2e^{0.5i})\right)\)
      • From the calculator: \(6e^{2i} + 2e^{0.5i} = -0.742 + 6.41i = 6.46e^{1.69i}\)
      • \(= \mathrm{Im}\left(6.46e^{(3t + 1.69)i}\right) = 6.46\sin(3t + 1.69)\)

  4. Write each expression in the form \(A\cos(x + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

    1. (a) \(12\cos x + 5\sin x\)

      • \(12\cos x + 5\sin x = \mathrm{Re}\left((12 - 5i)e^{xi}\right)\)
      • From the calculator: \(12 - 5i = 13e^{-0.395i}\)
      • \(= 13\cos(x - 0.395)\)

    2. (b) \(4\cos x - 7\sin x\)

      • \(4\cos x - 7\sin x = \mathrm{Re}\left((4 + 7i)e^{xi}\right)\)
      • From the calculator: \(4 + 7i = 8.06e^{1.05i}\)
      • \(= 8.06\cos(x + 1.05)\)

  5. The graph shows \(y = 4\cos x + 3\cos(x - 0.9)\) for \(0 \le x \le 6.5\). P is the maximum, Q is the minimum and R is the right-hand x-intercept.

    123456−9−6−3369xyPQR

    1. (a) Write \(y\) in the form \(A\cos(x + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(4\cos(x) + 3\cos(x - 0.9) = \mathrm{Re}\left(4e^{xi} + 3e^{(x - 0.9)i}\right)\)
      • \(= \mathrm{Re}\left(e^{xi}(4 + 3e^{-0.9i})\right)\)
      • From the calculator: \(4 + 3e^{-0.9i} = 5.86 - 2.35i = 6.32e^{-0.381i}\)
      • \(= \mathrm{Re}\left(6.32e^{(x - 0.381)i}\right) = 6.32\cos(x - 0.381)\)

    2. (b) Hence find the coordinates of P, Q and R.

      • \(P: x - 0.381 = 0 \Rightarrow x = 0.381,\ y = 6.32\)
      • \(Q: x - 0.381 = \pi \Rightarrow x = 3.52,\ y = -6.32\)
      • \(R: x - 0.381 = \tfrac{3\pi}{2} \Rightarrow x = 5.09,\ y = 0\)

  6. Two voltage sources feed one circuit. At time \(t\) seconds the voltages, in volts, are \(V_1 = 15\cos(40t)\) and \(V_2 = 10\cos(40t + 2)\).

    1. (a) Write the total voltage \(V(t) = V_1 + V_2\) in the form \(A\cos(40t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(15\cos(40t) + 10\cos(40t + 2) = \mathrm{Re}\left(15e^{40ti} + 10e^{(40t + 2)i}\right)\)
      • \(= \mathrm{Re}\left(e^{40ti}(15 + 10e^{2i})\right)\)
      • From the calculator: \(15 + 10e^{2i} = 10.8 + 9.09i = 14.1e^{0.698i}\)
      • \(= \mathrm{Re}\left(14.1e^{(40t + 0.698)i}\right) = 14.1\cos(40t + 0.698)\)

    2. (b) Find the greatest voltage.

      • \(\text{greatest } V = A = 14.1\) volts

In context

  1. A buoy floats near a harbour wall. At time \(t\) seconds it is lifted by waves from a ferry, \(H_1(t) = 0.5\sin(2t)\text{ m}\), and from a fishing boat, \(H_2(t) = 0.3\sin(2t + 1.5)\text{ m}\).

    1. (a) Write the total height \(H(t) = H_1 + H_2\) in the form \(A\sin(2t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(0.5\sin(2t) + 0.3\sin(2t + 1.5) = \mathrm{Im}\left(0.5e^{2ti} + 0.3e^{(2t + 1.5)i}\right)\)
      • \(= \mathrm{Im}\left(e^{2ti}(0.5 + 0.3e^{1.5i})\right)\)
      • From the calculator: \(0.5 + 0.3e^{1.5i} = 0.521 + 0.299i = 0.601e^{0.521i}\)
      • \(= \mathrm{Im}\left(0.601e^{(2t + 0.521)i}\right) = 0.601\sin(2t + 0.521)\)

    2. (b) Find the times in the first 10 seconds when \(H(t) = 0\).

      • \(0.601\sin(2t + 0.521) = 0 \Rightarrow 2t + 0.521 = k\pi\)
      • From the calculator, solving on \(0 \le t \le 10\): \(t = 1.31, 2.88, 4.45, 6.02, 7.59, 9.16\)

    3. (c) Find the first time at which the total height is greatest.

      • \(2t + 0.521 = \tfrac{\pi}{2} \text{ (smallest positive } t)\)
      • \(t = 0.525\)

    4. (d) At that time, what percentage of the wave is due to the ferry?

      • \(\dfrac{0.5\sin(2t)}{H(t)} \text{ at } t = 0.525: \dfrac{0.434}{0.601} \times 100 = 72.1\%\)

  2. The water level in a bay, measured from its mean level, is the sum of two tidal effects. At time \(t\) hours, in metres, \(D_1 = 1.8\cos(0.5t)\) and \(D_2 = 0.9\cos(0.5t + 2)\).

    1. (a) Write \(D(t) = D_1 + D_2\) in the form \(A\cos(0.5t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(1.8\cos(0.5t) + 0.9\cos(0.5t + 2) = \mathrm{Re}\left(1.8e^{0.5ti} + 0.9e^{(0.5t + 2)i}\right)\)
      • \(= \mathrm{Re}\left(e^{0.5ti}(1.8 + 0.9e^{2i})\right)\)
      • From the calculator: \(1.8 + 0.9e^{2i} = 1.43 + 0.818i = 1.64e^{0.521i}\)
      • \(= \mathrm{Re}\left(1.64e^{(0.5t + 0.521)i}\right) = 1.64\cos(0.5t + 0.521)\)

    2. (b) Find the greatest value of \(D\).

      • \(\text{greatest } D = A = 1.64\)

    3. (c) Find the first time \(t \ge 0\) at which \(D\) is greatest.

      • \(0.5t + 0.521 = 0 \Rightarrow t = 11.5\)

Stretch

  1. Three sound sources reach a microphone. At time \(t\) seconds the pressures, in pascals, are \(S_1 = 5\cos(2t)\), \(S_2 = 3\cos(2t + 1)\) and \(S_3 = 2\cos(2t - 2)\).

    1. (a) Write \(S(t) = S_1 + S_2 + S_3\) in the form \(A\cos(2t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(5\cos(2t) + 3\cos(2t + 1) + 2\cos(2t - 2) = \mathrm{Re}\left(5e^{2ti} + 3e^{(2t + 1)i} + 2e^{(2t - 2)i}\right) = \mathrm{Re}\left(e^{2ti}(5 + 3e^{1i} + 2e^{-2i})\right)\)
      • From the calculator: \(5 + 3e^{1i} + 2e^{-2i} = 5.79 + 0.706i = 5.83e^{0.121i}\)
      • \(= 5.83\cos(2t + 0.121)\)

    2. (b) State the greatest value of \(S\).

      • \(\text{greatest } S = A = 5.83\)

  2. Two radio signals add at a receiver. At time \(t\) milliseconds the strengths, in millivolts, are \(S_1 = 8\sin(7t)\) and \(S_2 = 5\sin(7t + 2)\).

    1. (a) Write \(R(t) = S_1 + S_2\) in the form \(A\sin(7t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(8\sin(7t) + 5\sin(7t + 2) = \mathrm{Im}\left(8e^{7ti} + 5e^{(7t + 2)i}\right)\)
      • \(= \mathrm{Im}\left(e^{7ti}(8 + 5e^{2i})\right)\)
      • From the calculator: \(8 + 5e^{2i} = 5.92 + 4.55i = 7.46e^{0.655i}\)
      • \(= \mathrm{Im}\left(7.46e^{(7t + 0.655)i}\right) = 7.46\sin(7t + 0.655)\)

    2. (b) At \(t = 0.1\), what percentage of the signal is due to \(S_1\)?

      • \(S_1(0.1) = 5.15,\ R(0.1) = 7.29\)
      • \(\dfrac{5.15}{7.29} \times 100 = 70.7\%\)

jmaths.xyz · AI HL · AHL 6.10 Adding sinusoidal functions with complex exponentials